Prove using the Triangle Inequality

[Chinnu]

Homework Statement

Show that:

(|x+y|)/(1+|x+y|) ≤ ((|x|)/(1+|x|)) + ((|y|)/(1+|y|))

Homework Equations

You are given the triangle inequality:

|x+y| ≤ |x| + |y|

The Attempt at a Solution

(This is done from the result, as I haven't been able to find the starting point)

(|x+y|)/(1+|x+y|) ≤ (|x|(1+|y|)+|y|(1+|x|))/((1+|x|)(1+|y|))

(|x+y|)/(1+|x+y|) ≤ (|x|+2|x||y|+|y|)/(1+|x|+|y|+|x||y|)

This doesn't seem to go anywhere. I also tried flipping the whole thing to get:

(1+|x+y|)/(|x+y|)≤(1+|x|)/(|x|)+(1+|y|)/(|y|)

but this doesn't seem to lead anywhere either...

I'm not sure how to go about this problem.

 

[spamiam]

As a first step, try starting by applying the triangle inequality to the numerator of the left-hand side. Next, what can you say about $1 + |x+y|$ and $1+|x|$? What does this imply about $\frac{1}{1+|x+y|}$ and $\frac{1}{1+|x|}$?

 

[Chinnu]

so, since 1+|x+y| $\geq$ 1+|x|,

$\frac{1}{1+|x+y|}$ $\leq$ $\frac{1}{1+|x|}$

Now,

$\frac{|x+y|}{1+|x+y|}$ $\leq$ $\frac{|x|}{1+|x+y|}$ + $\frac{|y|}{1+|x+y|}$

So,

$\frac{|x|}{1+|x+y|}$ + $\frac{|y|}{1+|x+y|}$ $\leq$ $\frac{|x|}{1+|x|}$ + $\frac{|y|}{1+|x+y|}$

Can a similar argument now simply be extended for 1+|y|?

 

[kru_]

You're on the right track. Prove the inequality is true for the denominator, then you can easily prove it is true for the numerator.

 

[spamiam]

so, since 1+|x+y| $\geq$ 1+|x|,

Actually, this isn't true: take x=1 and y=-1 for a counterexample.

I just noticed a trick that makes this much easier:

$$\frac{a}{1+a} = \frac{1+a-1}{1+a} = \frac{1+a}{1+a} - \frac{1}{1+a} = 1 - \frac{1}{1+a}\; .$$

Try using this trick on the LHS. Then you can use the triangle inequality (which will give you $1+|x+y| \leq 1 + |x| +|y|$) to compare $-\frac{1}{1+|x+y|}$ and $-\frac{1}{1+|x|+|y|}$. Then you can use the first trick backwards, which should lead to the result. And hopefully I haven't made any mistakes with my inequalities!